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Math Help - Equation of the tangent to a perpendicular lin

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    Equation of the tangent to a perpendicular lin

    Can someone help me to solve this : Find the equation of the tangent to the curve x2 -2xy + 2y2 -7x + 6y + 6 = 0 which is perpendicular to 6x + 5y -4 = 0
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    Re: Equation of the tangent to a perpendicular lin

    Quote Originally Posted by mgmanoj View Post
    Can someone help me to solve this : Find the equation of the tangent to the curve x2 -2xy + 2y2 -7x + 6y + 6 = 0 which is perpindendicular to 6x + 5y -4 = 0
    Use implicit differentiation to find y'.

    Find a point on the curve where y'=\frac{5}{6}.
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    Re: Equation of the tangent to a perpendicular lin

    After differentiating the equation I got
    dy = 2x-2y-7
    dx 2x-4y-6

    the slope of the tangent perpendicuar to 6x + 5y-4 = 0 is 45/6 or 5/6 - I got confused here.
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    Re: Equation of the tangent to a perpendicular lin

    Quote Originally Posted by mgmanoj View Post
    After differentiating the equation I got
    dy = 2x-2y-7
    dx 2x-4y-6

    the slope of the tangent perpendicuar to 6x + 5y-4 = 0 is 45/6 or 5/6 - I got confused here.
    You know \frac{2x-2y-7}{2x-4y-6}=\frac{5}{6}. Solve for either x\text{ or }y and substitute into the curve.
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    Re: Equation of the tangent to a perpendicular lin

    Quote Originally Posted by mgmanoj View Post
    After differentiating the equation I got
    dy = 2x-2y-7
    dx 2x-4y-6

    the slope of the tangent perpendicuar to 6x + 5y-4 = 0 is 45/6 or 5/6 - I got confused here.
    The equation 6x+ 5y- 4= 0 is the same as y= -(6/5)x+ 4/6 and has slope -6/5. The "negative reciprocal" of that is 6/5 and is the slope of the perpendicular line. I don't know where you got "45/6"! Did you mean "4/6"? Remember that the slope of the line, the number multiplying x, determines the direction, not the "constant" term.
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